Math Problem Statement

solve

The curve C has equation y = 2x^2 + px + q where p and q are integers. The curve C has a stationary point at (3, -5). (a) Show that p = -12 and find the value of q. (b) State, giving a reason, the nature of the stationary point. (c) Find an equation of the normal to C at the point on C where x = 1. Give your answer in the form ax + by + c = 0.

Solution

Let's solve the problem step by step. We are given the equation of a curve CC in the form:

y=2x2+px+qy = 2x^2 + px + q where pp and qq are integers. We also know that the curve has a stationary point at (3,5)(3, -5).

Part (a)

We are asked to show that p=12p = -12 and to find the value of qq.

Step 1: Differentiate the equation of the curve

To find the stationary points, we need to compute the derivative of yy with respect to xx. So, let's differentiate:

dydx=4x+p\frac{dy}{dx} = 4x + p

At a stationary point, the derivative is zero. We know the stationary point occurs at x=3x = 3, so substitute x=3x = 3 into the derivative:

4(3)+p=04(3) + p = 0 12+p=012 + p = 0 p=12p = -12

This confirms that p=12p = -12.

Step 2: Find qq

We are also given that the curve passes through the point (3,5)(3, -5). Substitute x=3x = 3, y=5y = -5, and p=12p = -12 into the equation of the curve:

5=2(3)2+(12)(3)+q-5 = 2(3)^2 + (-12)(3) + q 5=1836+q-5 = 18 - 36 + q 5=18+q-5 = -18 + q q=13q = 13

Thus, the value of qq is 1313.

Part (b)

We are asked to state, giving a reason, the nature of the stationary point.

Step 1: Second derivative test

To determine the nature of the stationary point, we use the second derivative. Differentiate dydx=4x12\frac{dy}{dx} = 4x - 12 again with respect to xx:

d2ydx2=4\frac{d^2y}{dx^2} = 4

Since d2ydx2=4\frac{d^2y}{dx^2} = 4, which is positive, the stationary point at (3,5)(3, -5) is a minimum.

Part (c)

We are asked to find the equation of the normal to the curve at the point where x=1x = 1.

Step 1: Find the gradient of the tangent at x=1x = 1

The gradient of the tangent to the curve is given by the first derivative, dydx=4x12\frac{dy}{dx} = 4x - 12. Substituting x=1x = 1:

dydx=4(1)12=412=8\frac{dy}{dx} = 4(1) - 12 = 4 - 12 = -8

The gradient of the tangent is 8-8, so the gradient of the normal is the negative reciprocal of this:

Gradient of the normal=18\text{Gradient of the normal} = \frac{1}{8}

Step 2: Find the coordinates of the point on the curve where x=1x = 1

Substitute x=1x = 1 into the equation of the curve to find the yy-coordinate:

y=2(1)212(1)+13y = 2(1)^2 - 12(1) + 13 y=212+13=3y = 2 - 12 + 13 = 3

So, the point on the curve is (1,3)(1, 3).

Step 3: Equation of the normal

The equation of a line is given by:

yy1=m(xx1)y - y_1 = m(x - x_1)

where mm is the gradient of the normal, and (x1,y1)(x_1, y_1) is the point (1,3)(1, 3). Substituting the values:

y3=18(x1)y - 3 = \frac{1}{8}(x - 1)

Multiply through by 8 to eliminate the fraction:

8(y3)=x18(y - 3) = x - 1 8y24=x18y - 24 = x - 1 x8y+23=0x - 8y + 23 = 0

Thus, the equation of the normal is:

x8y+23=0x - 8y + 23 = 0


Summary of answers:

  • (a) p=12p = -12, q=13q = 13
  • (b) The stationary point is a minimum because d2ydx2=4\frac{d^2y}{dx^2} = 4, which is positive.
  • (c) The equation of the normal is x8y+23=0x - 8y + 23 = 0.

Do you have any questions or need further details?

Related Questions:

  1. How do you find the second derivative of more complex functions?
  2. What does the second derivative tell us about a curve's shape?
  3. Can there be stationary points that are neither maxima nor minima?
  4. How do we find the tangent to a curve at a given point?
  5. What role does the discriminant play in determining the nature of stationary points?

Tip: Always check if the second derivative is positive or negative at a stationary point to confirm whether it is a minimum or maximum.

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Math Problem Analysis

Mathematical Concepts

Calculus
Quadratic Functions
Derivatives
Stationary Points

Formulas

Derivative of y = 2x^2 + px + q is dy/dx = 4x + p
Second derivative: d^2y/dx^2 = 4
Equation of a line: y - y1 = m(x - x1)

Theorems

Second Derivative Test

Suitable Grade Level

Grades 10-12